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Oracle Java SE 21 Developer Professional Sample Questions (Q50-Q55):

NEW QUESTION # 50
What do the following print?
java
import java.time.Duration;
public class DividedDuration {
public static void main(String[] args) {
var day = Duration.ofDays(2);
System.out.print(day.dividedBy(8));
}
}

A. Compilation fails
B. It throws an exception
C. PT6H
D. PT0D
E. PT0H

Answer: C

Explanation:
In this code, a Duration object day is created representing a duration of 2 days using the Duration.ofDays(2) method. The dividedBy(long divisor) method is then called on this Duration object with the argument 8.
The dividedBy(long divisor) method returns a copy of the original Duration divided by the specified value. In this case, dividing 2 days by 8 results in a duration of 0.25 days. In the ISO-8601 duration format used by Java's Duration class, this is represented as PT6H, which stands for a period of 6 hours.
Therefore, the output of the System.out.print statement is PT6H.

NEW QUESTION # 51
Which methods compile?

A. ```java public List<? extends IOException> getListExtends() { return new ArrayList<Exception>(); } csharp
B. ```java
public List<? extends IOException> getListExtends() {
return new ArrayList<FileNotFoundException>();
}
C. ```java public List<? super IOException> getListSuper() { return new ArrayList<Exception>(); } csharp
D. ```java
public List<? super IOException> getListSuper() {
return new ArrayList<FileNotFoundException>();
}

Answer: B,C

Explanation:
In Java generics, wildcards are used to relax the type constraints of generic types. The extends wildcard (<?
extends Type>) denotes an upper bounded wildcard, allowing any type that is a subclass of Type. Conversely, the super wildcard (<? super Type>) denotes a lower bounded wildcard, allowing any type that is a superclass of Type.
Option A:
java
public List<? super IOException> getListSuper() {
return new ArrayList<Exception>();
}
Here, List<? super IOException> represents a list that can hold IOException objects and objects of its supertypes. Since Exception is a superclass of IOException, ArrayList<Exception> is compatible with List<?
super IOException>. Therefore, this method compiles successfully.
Option B:
java
public List<? extends IOException> getListExtends() {
return new ArrayList<FileNotFoundException>();
}
In this case, List<? extends IOException> represents a list that can hold objects of IOException and its subclasses. Since FileNotFoundException is a subclass of IOException, ArrayList<FileNotFoundException> is compatible with List<? extends IOException>. Thus, this method compiles successfully.
Option C:
java
public List<? extends IOException> getListExtends() {
return new ArrayList<Exception>();
}
Here, List<? extends IOException> expects a list of IOException or its subclasses. However, Exception is a superclass of IOException, not a subclass. Therefore, ArrayList<Exception> is not compatible with List<?
extends IOException>, and this method will not compile.
Option D:
java
public List<? super IOException> getListSuper() {
return new ArrayList<FileNotFoundException>();
}
In this scenario, List<? super IOException> expects a list that can hold IOException objects and objects of its supertypes. Since FileNotFoundException is a subclass of IOException, ArrayList<FileNotFoundException> is not compatible with List<? super IOException>, and this method will not compile.
Therefore, the methods in options A and B compile successfully, while those in options C and D do not.

NEW QUESTION # 52
Given:
java
StringBuffer us = new StringBuffer("US");
StringBuffer uk = new StringBuffer("UK");
Stream<StringBuffer> stream = Stream.of(us, uk);
String output = stream.collect(Collectors.joining("-", "=", ""));
System.out.println(output);
What is the given code fragment's output?

A. US=UK
B. Compilation fails.
C. US-UK
D. An exception is thrown.
E. =US-UK
F. -US=UK

Answer: E

Explanation:
In this code, two StringBuffer objects, us and uk, are created with the values "US" and "UK", respectively. A stream is then created from these objects using Stream.of(us, uk).
The collect method is used with Collectors.joining("-", "=", ""). The joining collector concatenates the elements of the stream into a single String with the following parameters:
* Delimiter ("-"):Inserted between each element.
* Prefix ("="):Inserted at the beginning of the result.
* Suffix (""):Inserted at the end of the result.
Therefore, the elements "US" and "UK" are concatenated with "-" between them, resulting in "US-UK". The prefix "=" is added at the beginning, resulting in the final output =US-UK.

NEW QUESTION # 53
Given:
java
import java.io.*;
class A implements Serializable {
int number = 1;
}
class B implements Serializable {
int number = 2;
}
public class Test {
public static void main(String[] args) throws Exception {
File file = new File("o.ser");
A a = new A();
var oos = new ObjectOutputStream(new FileOutputStream(file));
oos.writeObject(a);
oos.close();
var ois = new ObjectInputStream(new FileInputStream(file));
B b = (B) ois.readObject();
ois.close();
System.out.println(b.number);
}
}
What is the given program's output?

A. 0
B. 1
C. Compilation fails
D. NotSerializableException
E. ClassCastException

Answer: E

Explanation:
In this program, we have two classes, A and B, both implementing the Serializable interface, and a Test class with the main method.
Program Flow:
* Serialization:
* An instance of class A is created and assigned to the variable a.
* An ObjectOutputStream is created to write to the file "o.ser".
* The object a is serialized and written to the file.
* The ObjectOutputStream is closed.
* Deserialization:
* An ObjectInputStream is created to read from the file "o.ser".
* The program attempts to read an object from the file and cast it to an instance of class B.
* The ObjectInputStream is closed.
Analysis:
* Serialization Process:
* The object a is an instance of class A and is serialized into the file "o.ser".
* Deserialization Process:
* When deserializing, the program reads the object from the file and attempts to cast it to class B.
* However, the object in the file is of type A, not B.
* Since A and B are distinct classes with no inheritance relationship, casting an A instance to B is invalid.
Exception Details:
* Attempting to cast an object of type A to type B results in a ClassCastException.
* The exception message would be similar to:
pgsql
Exception in thread "main" java.lang.ClassCastException: class A cannot be cast to class B Conclusion:
The program compiles successfully but throws a ClassCastException at runtime when it attempts to cast the deserialized object to class B.

NEW QUESTION # 54
Given:
java
int post = 5;
int pre = 5;
int postResult = post++ + 10;
int preResult = ++pre + 10;
System.out.println("postResult: " + postResult +
", preResult: " + preResult +
", Final value of post: " + post +
", Final value of pre: " + pre);
What is printed?

A. postResult: 15, preResult: 16, Final value of post: 6, Final value of pre: 6
B. postResult: 16, preResult: 15, Final value of post: 6, Final value of pre: 5
C. postResult: 16, preResult: 16, Final value of post: 6, Final value of pre: 6
D. postResult: 15, preResult: 16, Final value of post: 5, Final value of pre: 6

Answer: A

Explanation:
* Understanding post++ (Post-increment)
* post++uses the value first, then increments it.
* postResult = post++ + 10;
* post starts as 5.
* post++ returns 5, then post is incremented to 6.
* postResult = 5 + 10 = 15.
* Final value of post after this line is 6.
* Understanding ++pre (Pre-increment)
* ++preincrements the value first, then uses it.
* preResult = ++pre + 10;
* pre starts as 5.
* ++pre increments pre to 6, then returns 6.
* preResult = 6 + 10 = 16.
* Final value of pre after this line is 6.
Thus, the final output is:
yaml
postResult: 15, preResult: 16, Final value of post: 6, Final value of pre: 6 References:
* Java SE 21 - Operators and Expressions
* Java SE 21 - Arithmetic Operators

NEW QUESTION # 55
......

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